3 ms·

A pdf at [0, N] sorts before the one at [0, N+1], by "first valid pdf".

by infogulch 6y ago

paulmd 6y ago
No, both start at 0. Also, [0, infinity] and [0, infinity+1] are the same thing.
- infogulch 6y ago
  https://www.online-python.com/OuZ0thRs6D https://www.online-python.com/OuZ0thRs6D
  - paulmd 6y ago
    your example fails to satisfy the invariant. 11 is less than infinity. you're just pasting random python snippits at me now. It's time to move on. again, just to summarize: PDF files do not have to be zero aligned, and they do not have to be end aligned. Therefore the answer to the question "what is the first segment of Pi that is a valid PDF file" is trivially (0,infinity). That is a correct statement. The non-greedy (in the regex sense) answer to that question will be different, however.
    - infogulch 6y ago
      Why is this so hard? If the tuple (0,10) represents the range of a valid pdf, then the next tuple (0,11) is also a valid pdf. Or any after it up to and including (0,infinity). Note the word "next", implying that (0,10) sorts before (0,11); you even say it yourself "11 is less than infinity". Where I'm from "first" and "less" are related (the first element in a unique sorted list is defined to be less than all other elements). So if there is any valid pdf in pi that can be identified by the range tuple (0,N), then the first valid pdf must occur before N -> infinity. Therefore (0,infinity) can never be the first valid pdf, even though it may be a valid pdf. Maybe a picture would help: Potential pdf file ranges in pi: [(0,0),(0,1),(0,2),(0,3),(0,4),...,(0,N-1),(0,N),(0,N+1),(0,N+2),...,(0,infinity)] Is it a valid pdf? no no no no no (no) no yes yes yes (yes) yes Which one is first? ^^^ I thought linking to a python script that shows the order comparison of a tuple (0,N) as less than the tuple (0,N+1) would clearly demonstrate this, but it appears to have failed to communicate that to you. We don't need non-greedy regex rules to do a less than comparison.

A pdf at [0, N] sorts before the one at [0, N+1], by "first valid pdf".

paulmd 6y ago

No, both start at 0. Also, [0, infinity] and [0, infinity+1] are the same thing.

infogulch 6y ago
https://www.online-python.com/OuZ0thRs6D https://www.online-python.com/OuZ0thRs6D
- paulmd 6y ago
  your example fails to satisfy the invariant. 11 is less than infinity. you're just pasting random python snippits at me now. It's time to move on. again, just to summarize: PDF files do not have to be zero aligned, and they do not have to be end aligned. Therefore the answer to the question "what is the first segment of Pi that is a valid PDF file" is trivially (0,infinity). That is a correct statement. The non-greedy (in the regex sense) answer to that question will be different, however.
  - infogulch 6y ago
    Why is this so hard? If the tuple (0,10) represents the range of a valid pdf, then the next tuple (0,11) is also a valid pdf. Or any after it up to and including (0,infinity). Note the word "next", implying that (0,10) sorts before (0,11); you even say it yourself "11 is less than infinity". Where I'm from "first" and "less" are related (the first element in a unique sorted list is defined to be less than all other elements). So if there is any valid pdf in pi that can be identified by the range tuple (0,N), then the first valid pdf must occur before N -> infinity. Therefore (0,infinity) can never be the first valid pdf, even though it may be a valid pdf. Maybe a picture would help: Potential pdf file ranges in pi: [(0,0),(0,1),(0,2),(0,3),(0,4),...,(0,N-1),(0,N),(0,N+1),(0,N+2),...,(0,infinity)] Is it a valid pdf? no no no no no (no) no yes yes yes (yes) yes Which one is first? ^^^ I thought linking to a python script that shows the order comparison of a tuple (0,N) as less than the tuple (0,N+1) would clearly demonstrate this, but it appears to have failed to communicate that to you. We don't need non-greedy regex rules to do a less than comparison.